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The question is a bit weird since the wave function is quantum mechanical and the distribution function in phase space is really something classical.

But I would still like to know if I take the square of the wave function $$\langle x|p\rangle =e^{-\text{ip}\cdot x},$$ that means I would have an equal probability of finding particles in positional space.

Now if I scan all the states $|p\rangle$, and assume that all momentum states in representation $\langle x|$ are plane waves, then the corresponding probability for finding the particle would be equal in positional space for all $|p\rangle$.

Can I conclude now that in this case, I will have a distribution function $f(x,p)=constant$?

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The concept of phase-space distribution is problematic in quantum mechanics. As I am sure you know, the Heisenberg uncertainty principle restricts the widths of the position and momentum distributions through

$$\Delta x\Delta p\geq\dfrac{\hbar}{2}$$

so the notion of $f\left(x,p\right)$ as the probability of a particle being at the exact point $x$ with exact momentum $p$ is ill-defined.

That being said, there is a way to get functions known as quasi-probability distributions that posses some of the properties of classical distribution. One example of such a function is the Wigner quasi-probability distribution defined as

$$W\left(x,p\right)=\frac{1}{\pi\hbar}\int_{-\infty}^{\infty}\psi^{\ast}\left(x+y\right)\psi\left(x-y\right)e^{2ipy/\hbar}{\rm d}y$$

It is similar to a probability density, but in some cases can admit negative values (which is highly non-classical behavior). In the case of a wavefunction of a particle with definite momentum $\hbar k$

$$\psi\left(x\right)=\dfrac{1}{\sqrt{2\pi}}e^{ikx}$$

the Wigner distribution has the form

$$W\left(x,p\right)=\frac{1}{\pi\hbar}\int_{-\infty}^{\infty}\dfrac{1}{\sqrt{2\pi}}e^{-ik\left(x+y\right)}\dfrac{1}{\sqrt{2\pi}}e^{ik\left(x-y\right)}e^{2ipy/\hbar}{\rm d}y= \\\\ =\frac{1}{2\pi^{2}\hbar}\int_{-\infty}^{\infty}e^{2iy\left(\frac{p}{\hbar}-k\right)}{\rm d}y=\frac{1}{4\pi^{2}}\int_{-\infty}^{\infty}e^{iy\left(p-\hbar k\right)}{\rm d}y=\dfrac{1}{2\pi}\delta\left(p-\hbar k\right)$$

For a classical particle of mass $m$ with the same momentum $q=\hbar k$ you would have a different distribution

$$f\left(x,p\right)=\delta\left(x-\dfrac{q}{m}t\right)\delta\left(p-q\right)$$

both of which are not constants.

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